nicole_柠檬草
幼苗
共回答了21个问题采纳率:90.5% 举报
y=x+16/(x+2),x≥4
设4≤x1<x2
故:x1+2≥6,x2+2>x1+2≥6,x2-x1>0
故:(x1+2)(x2+2) >36
故:0<16/[(x1+2)(x2+2)] <1
故:1-16/[(x1+2)(x2+2)]>0
故:f(x2)-f(x1)= x2+16/(x2+2)-x1-16/(x1+2)
=(x2-x1){1-16/[(x1+2)(x2+2)]} >0
故:x≥4时,y=f(x)=x+16/(x+2)单调递增
即:x=4时,y=f(x)=x+16/(x+2)取最小值,最小值为20/3
故:y≥20/3
1年前
2